Six point charges are kept $$60^{o}$$ apart from each other on the circumference of a
circle of radius $$R$$ as shown in figure . The net electric field at the center of the circle is______.
( $$\epsilon_{o}$$ is permittivity of free space)

Using components:

Magnitude due to each charge,

$$E_0=\frac{1}{4\pi\ \epsilon_o}.\left(\frac{Q}{R^2}\right)$$

Fields from top and bottom charges cancel.

Now remaining four charges:

At $$30^{\circ}$$,

$$\vec{E}_1=-E_0\left(\frac{\sqrt{3}}{2}i+\frac{1}{2}j\right)$$

At $$-30^{\circ}$$,

$$\vec{E}_2=-E_0\left(\frac{\sqrt{3}}{2}i-\frac{1}{2}j\right)$$

At $$150^{\circ}$$ (negative charge),

$$\vec{E}_3=E_0\left(-\frac{\sqrt{3}}{2}i+\frac{1}{2}j\right)$$

At $$-150^{\circ}$$,

$$\vec{E}_4=E_0\left(\frac{\sqrt{3}}{2}i+\frac{1}{2}j\right)$$

Adding x-components:

$$E_x=-\frac{\sqrt{3}}{2}E_0-\frac{\sqrt{3}}{2}E_0-\frac{\sqrt{3}}{2}E_0+\frac{\sqrt{3}}{2}E_0$$

$$=-\sqrt{3}E_0$$

Adding y-components:

$$E_y=-\frac{1}{2}E_0+\frac{1}{2}E_0+\frac{1}{2}E_0+\frac{1}{2}E_0$$

=$$E_0$$

Therefore,

$$E=-\sqrt{\ 3}E_0i\ \ +\ E_0j$$

Substitute $$E_0$$​:

$$-\frac{Q}{4\pi\epsilon_{o}R^{2}}\left(\sqrt{3}\widehat{i}- \widehat{j} \right)$$